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trig problem

Discussion in 'Chit Chat' started by Jakeman, May 7, 2003.

  1. Jakeman

    Jakeman MSC Founder and Donator

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    Directions:
    Find all solutions in the interval (0, 2pi]

    Problem:
    tan^2(x)*csc(x) = tan^2(x)

    Translated into English since it might look confusing typed out... the tangeant squared of x times the cosecant of x equals the tangeant squared of x. That is not the tangeant of (x^2).

    I determined there were no solutions, but that wasn't the right answer. I still think I got it right. If I can get the points for this one problem, I will have an A instead of a B in that class for this past Spring semester.

    Can you guys tell me what answer you get and how you worked it out?
     
  2. Jakeman

    Jakeman MSC Founder and Donator

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  3. kraahl

    kraahl Peasant

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    the thing that confuses me is csc(x), I´m only familiar with tan(x), cos(x) and sin(x)
    or perhaps we write it differnently, but I doubt that.

    well, physics final tomorrow... I´d better go study
     
  4. Jakeman

    Jakeman MSC Founder and Donator

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    csc is just 1/sin
     
  5. OsIriS

    OsIriS Peasant

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    jakeman just spelled it wrong :)...I would have been able to do that problem for you a LONG time ago, but I forgot all my trig..isn't tangeant, cosxsin or something like that? I seem to remember you have to turn them all into like terms or something...but my brain explodes when thinking that far back ('94 guys, almost 10 years.....ouch!!!! *bonk*
     
  6. OsIriS

    OsIriS Peasant

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    whoops.....:bonk:
     
  7. aleldtritch

    aleldtritch Peasant

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    Well since the first function times another function equals the first function, the second function must be 1.

    tan^2(x)*csc(x)=tan^2(x)
    -I know this isn't technically allowed, but divide both sides by tan^2(x)
    csc(x)=1

    so 0 and 2pi

    edit- 0 and 2pi is wrong, I mixed up csc with sec...it's pi/2

    edit2-oh yeah, tan is undefined with that. There's no answer.
     
    Last edited: May 7, 2003
  8. kraahl

    kraahl Peasant

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    From that Chen guys post in the other forum

    "sin(x) = sin^2(x)
    Or:
    sin(x) * [1 - sin(x)] = 0"

    wouldn´t it be

    sin(x) = 1- cos^2(x)
     
  9. Jakeman

    Jakeman MSC Founder and Donator

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    That's the exact logic I used. I got to an answer of pi/2, but that made tan undefined so I said there was no answer. But I neglected to consider the possible solution of tan(x) equalling 0.

    At 0 and pi, the equation equates to 0 * undefined = 0 . I guess "undefined" is legit when it's multiplied by 0.
     
  10. mrdoomsday

    mrdoomsday Peasant

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    well if its basicaly
    a*b=a

    couldnt you work your way backwards using the negative function?

    though im not familiar with cosecant. and i cant find a button familiar to csc on ti-83.. Is this some weird form of cosign?
     
  11. Jakeman

    Jakeman MSC Founder and Donator

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    csc = 1/sin

    So calculate the sin, then hit your inverse key. Or store the value, clear, 1 divided by *recall value*, enter.
     
  12. Haite

    Haite Forum Moderator

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    It's the sin^-1 key on your calculator.

    Consider replacing tangent in the problem with its definition (sin / cos).

    My solution is X = π / 2
     
  13. Jakeman

    Jakeman MSC Founder and Donator

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    The tangeant is undefined at pi/2

    sin(pi/2) = 1
    cos(pi/2) = 0
    tan(x) = sin(x)/cos(x)
    tan(pi/2) = sin(pi/2)/cos(pi/2)
    tan(pi/2) = 1/0
    tan(pi/2) = undefined

    Be careful when multiplying or dividing both sides of an equation by a variable quantity. You can introduce extraneous solutions that must be checked in the original equation at the end.
     
  14. cowofwar

    cowofwar Peasant

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    I hate that shit, it's easy, but I still hate it.
     
  15. mud

    mud Peasant

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    Not quite related to the problem, but I have my own problem, i need to find the derivitive of y = e^e^x
     
  16. cowofwar

    cowofwar Peasant

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    ln both sides to get
    lny = lne^e^x
    lne = 1
    so
    lny = e^x
    dy/dx (lny) = dy/dx (e^x)
    dy/dx (1/y) = e^x

    NIGGA PLZ
     
    Last edited: May 8, 2003
  17. mud

    mud Peasant

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    Umm, okay, except my book says e^(x+e^x)

    I have no clue how they got that answer.
     
  18. cowofwar

    cowofwar Peasant

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    Ask your prof, I don't see where that could come from unless they provide you with conditions to solve.
     
  19. cowofwar

    cowofwar Peasant

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    Lolz. Nevermind.

    They're not solving with ln, just solving with basics.

    So you have y = e^(e^x)

    Power rule, multiply the original part by the derivative of the exponent. The derivative of e^x is e^x so just multiple e^e^x by e^x and you end up with

    y = (e^x)(e^e^x)
    Simplify
    y = e^(x+e^x)
    by exponent laws.


    Seems weird to me.
     
  20. aleldtritch

    aleldtritch Peasant

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    AP Calc exam here I come. I hope there aren't many revolution of solids questions.
     

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